Java: Prefix/postfix of increment/decrement operators?

Question

From the program below or here, why does the last call to System.out.println(i) print the value 7?

class PrePostDemo {
     public          


        

Answer 1

Why wouldn't the variable have been updated?

• Postfix: passes the current value of i to the function and then increments it.
• Prefix: increments the current value and then passes it to the function.

The lines where you don't do anything with i make no difference.

Notice that this is also true for assignments:

i = 0;
test = ++i;  // 1
test2 = i++; // 1

Answer 2

It prints 7 for the last statement, cos in the statement above, it's value is 6 and it's incremented to 7 when the last statement gets printed

Answer 3

This is my answer. Some of you may find it easy to understand.

package package02;

public class C11PostfixAndPrefix {

    public static void main(String[] args) {
        // In this program, we will use the value of x for understanding prefix 
        // and the value of y for understaning postfix. 
        // Let's see how it works. 

        int x = 5; 
        int y = 5; 

        Line 13:   System.out.println(++x);  // 6   This is prefixing. 1 is added before x is used. 
        Line 14:   System.out.println(y++);  // 5   This is postfixing. y is used first and 1 is added. 

        System.out.println("---------- just for differentiating");

        System.out.println(x);  // 6   In prefixing, the value is same as before {See line 13}
        System.out.println(y);  // 6   In postfixing, the value increases by 1  {See line 14} 

        // Conclusion: In prefixing (++x), the value of x gets increased first and the used 
        // in an operation. While, in postfixing (y++), the value is used first and changed by
        // adding the number. 
    }
}

Answer 4

Well think of it in terms of temporary variables.

i =3 ;
i ++ ; // is equivalent to:   temp = i++; and so , temp = 3 and then "i" will increment and become     i = 4;
System.out.println(i); // will print 4

Now,

i=3;
System.out.println(i++);

is equivalent to

temp = i++;  // temp will assume value of current "i", after which "i" will increment and become i= 4
System.out.println(temp); //we're printing temp and not "i"